  Question

A train travels at a certain average speed of a distance of $$63\ km$$ and then travels at a distance of $$72\ km$$ at an average speed of $$6\ km/hr$$   more than its original speeds. If it takes $$3$$ hours to complete total journey, what is the original average speed?

Solution

Let, the original average speed be $$x$$ km/hr.Time taken at a speed of $$x$$ km/hr $$(t_1)=\dfrac{63}{x}$$ hrsTime taken at a speed of $$x+6$$ km/hr $$(t_2)=\dfrac{72}{x+6}$$ hrs$$t_1+t_2=3$$$$\dfrac{63}{x}+\dfrac{72}{x+6}=3$$$$\implies \dfrac{63((x+6)+72x}{x(x+6)}=3$$$$\implies 63x+378+72x=3x(x+6)$$$$\implies 378+135x=3x^2+18x$$$$\implies 3x^2+18x-135x-378=0$$$$\implies 3x^2-117x-378=0$$$$\implies 3(x^2-39x-126)=0$$$$\implies x^2-39x-126=0$$$$\implies x^2-42x+3x-126=0$$$$\implies x(x-42)+3(x-42)=0$$$$\implies (x-42)(x+3)=0$$Either $$x=42$$ or $$x=-3$$$$\because$$ speed cannot be negative. $$x=42$$ is considered.$$\therefore$$ The original average speed is $$42$$ km/hrPhysics

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